Dynamic price dispersion in Bertrand–Edgeworth competition

Abstract

This paper studies a dynamic oligopoly model of price competition under demand uncertainty. Sellers are endowed with one unit of the good and compete by posting prices in every period. Buyers each demand one unit of the good and have a common reservation price. They have full information regarding the prices posted by each firm in the market; hence, search is costless. The number of buyers coming to the market in each period is random. Demand uncertainty is said to be high if there are at least two non-zero demand states that give a seller different option values of waiting to sell. Our model features a unique symmetric Markov perfect equilibrium in which price dispersion prevails if and only if the degree of demand uncertainty is high. Several testable theoretical implications on the distribution of market prices are derived.

Appendix

Proof of Proposition  1

Pick any period $t\in \mathbb {N} .$ If in period t there is only one seller remaining in the market, then the seller will simply set a price $p=\overline{p}=p_{1}^{*}$ in every period. Suppose there are two sellers in the market in period t. Let $F_{i}(p):[0,\infty )\rightarrow [0,1]$ denote the strategy played by seller i in period t ($F_{i}(p)$ could be degenerate) in an equilibrium, $i=1,2$. Define $l_{i}=\sup \{p:F_{i}(p)=0\}$ and $u_{i}=\inf \{p:F_{i}(p)=1\}.$ Define $V_{i}(t)$ to be the present discounted value of seller $i^{\prime }s$ expected profit in period t. Conditional on demand being zero in period t (no buyer shows up in period t), $V_{i}(t+1)=V_{i}(t)$. We prove the proposition by establishing the following claims:

Claim 1. $l_{1}=l_{2}\ge p_{2}^{*}$: Suppose $l_{1}\ne l_{2}$ and WLOG let $l_{1}<l_{2}.$ Seller 1 can increase his profit by putting all probability between $[l_{1},l_{2})$ to $l_{2}-\varepsilon .$ Hence in any equilibrium we must have $l_{1}=l_{2}.$ Suppose now we have $l_{1}=l_{2}<p_{2}^{*}.$ Consider two cases:

Case i. $F_{2}(l_{2})<1.$ Since a distribution is right continuous, $\exists$ $\eta \in (l_{2},p_{2}^{*})$ such that $F_{2}(l_{2})\le F_{2}(p)<1$ $\forall$ $p\in (l_{2},\eta ).$ Given seller 2’s equilibrium strategy, seller 1’s expected profit from setting a price $p\in [l_{2},\eta )$ is strictly lower than that from posting the price at $p=p_{2}^{*}$:¹³

$$\begin{aligned} V_{1}(t)&=q\delta V_{1}(t)+(1-q)\left[ \phantom {\left. +(F_{2}(p)-F_{2}(p-))\left( \frac{1}{2}\delta V(\overline{p},1)+\frac{1}{2} p\right) +(1-F_{2}(p))p\right] }F_{2}(p-)\delta V(\overline{p} ,1)\right. \nonumber \\&\left. +\,(F_{2}(p)-F_{2}(p-))\left( \frac{1}{2}\delta V(\overline{p},1)+\frac{1}{2} p\right) +(1-F_{2}(p))p\right] \nonumber \\&<q\delta V_{1}(t)+(1-q)\left[ \phantom {\left. +(F_{2}(p)-F_{2}(p-))\delta V(\overline{p},1)+(1-F_{2}(p))\delta V(\overline{p},1)\right] }F_{2}(p-)\delta V(\overline{p} ,1)\right. \nonumber \\&\left. +\,(F_{2}(p)-F_{2}(p-))\delta V(\overline{p},1)+(1-F_{2}(p))\delta V(\overline{p},1)\right] \nonumber \\&=q\delta V_{1}(t)+(1-q)\left[ F_{2}(p_{2}^{*}-)\delta V(\overline{p} ,1)\right. \nonumber \\&\left. +\,(F_{2}(p_{2}^{*})-F_{2}(p_{2}^{*}-))\frac{1}{2}(\delta V(\overline{p},1)+p_{2}^{*}) \right. \nonumber \\&\left. +\,(1-F_{2}(p_{2}^{*}))p_{2}^{*}\right] \end{aligned}$$

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Here we use the fact that $p_{2}^{*}=\delta V(\overline{p},1)>\eta >p.$ Consequently, seller 1 will not post any price $p\in [l_{2},\eta )$ with positive probability,  and we have $l_{1}=\sup \{p:F_{1}(p)=0\}\ge \eta ,$ which contradicts the premise that $l_{1}=l_{2}<\eta$.

Case ii. $F_{2}(l_{2})=1.$ $F_{2}(l_{2})=1$ indicates that seller 2 sets the price at $l_{2}$ with probability 1. Given seller 1’s equilibrium strategy, seller 2’s expected profit from posting the price at $p=l_{2}$ is strictly lower than that from posting the price at $p=p_{2}^{*}:$

$$\begin{aligned} V_{2}(t)= & {} q\delta V_{2}(t)+(1-q)\left[ F_{1}(l_{2})\left[ \frac{1}{2}\delta V( \overline{p},1)+\frac{1}{2}l_{2}\right] +(1-F_{1}(l_{2}))l_{2}\right] \nonumber \\< & {} q\delta V_{2}(t)+(1-q)\delta V(\overline{p},1) \nonumber \\= & {} q\delta V_{2}(t)+(1-q)\left[ F_{1}(p_{2}^{*}-)\delta V(\overline{p} ,1)\right. \nonumber \\&\left. +\,(F_{1}(p_{2}^{*})-F_{1}(p_{2}^{*}-))\frac{1}{2}(\delta V( \overline{p},1)+p_{2}^{*}) \right. \nonumber \\&\left. +\,(1-F_{1}(p_{2}^{*}))p_{2}^{*}\right] , \end{aligned}$$

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hence a contradiction is reached.

We have shown that both cases are not feasible. Thus in an equilibrium we must have $l_{1}=l_{2}\ge p_{2}^{*}.$

Claim 2. $u_{1}=u_{2}=p_{2}^{*}$: Suppose $u_{1}\ne u_{2}$ and WLOG let $u_{1}<u_{2}.$ By Claim 1, $p_{2}^{*}\le u_{1}<u_{2}.$ First, we observe that $p_{2}^{*}<u_{1}<u_{2}$ cannot be supported by any equilibrium, for if it is indeed the case, then seller 2’s expected profit from posting any price $p\in (p_{2}^{*},u_{1})$ is greater than that from posting any price $p>u_{1},$ $i.e.,\forall$ $p\in (p_{2}^{*},u_{1})$

$$\begin{aligned} V_{2}(t)= & {} q\delta V_{2}(t)+(1-q)\left[ \phantom {\left. +(F_{1}(p)-F_{1}(p-))\left( \frac{1}{2}\delta V(\overline{p},1)+\frac{1}{2} p\right) +(1-F_{1}(p))p\right] }F_{1}(p-)\delta V(\overline{p} ,1)\right. \nonumber \\&\left. +\,(F_{1}(p)-F_{1}(p-))\left( \frac{1}{2}\delta V(\overline{p},1)+\frac{1}{2} p\right) +(1-F_{1}(p))p\right] \nonumber \\> & {} q\delta V_{2}(t)+(1-q)[F_{1}(p-)\delta V(\overline{p} ,1)\nonumber \\&+\,(F_{1}(p)-F_{1}(p-))\delta V(\overline{p},1)+(1-F_{1}(p))\delta V( \overline{p},1)] \nonumber \\= & {} q\delta V_{2}(t)+(1-q)\delta V(\overline{p},1), \end{aligned}$$

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which implies $\inf \{p:F_{2}(p)=1\}\le u_{1},$ a contradiction. On the other hand, $p_{2}^{*}=u_{1}<u_{2}$ also cannot be sustained in any equilibrium, for seller 1’s expected profit from posting any price $p\in (p_{2}^{*},u_{2})$ is greater than that from posting any price $p\le p_{2}^{*},$ i.e., $\forall$ $p\in (p_{2}^{*},u_{2})$

$$\begin{aligned} V_{1}(t)&=q\delta V_{1}(t)+(1-q)\left[ \phantom {\left. \left( \frac{1}{2}\delta V(\overline{p},1)+\frac{1}{2} p\right) +(1-F_{2}(p))p\right] }F_{2}(p-)\delta V(\overline{p} ,1)+(F_{2}(p)-F_{2}(p-))\right. \nonumber \\&\quad \left. \left( \frac{1}{2}\delta V(\overline{p},1)+\frac{1}{2} p\right) +(1-F_{2}(p))p\right] \nonumber \\&>q\delta V_{1}(t)+(1-q)\delta V(\overline{p},1), \end{aligned}$$

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which implies that $F_{1}(p_{2}^{*})=0,$ violating the premise that $u_{1}=\inf \{p:F_{1}(p)=1\}=p_{2}^{*}.$ Therefore we have shown that $u_{1}=u_{2}$ in any equilibrium.

Now suppose $u_{1}=u_{2}>p_{2}^{*}.$ Combining the result from Claim 1, there are two possible cases to be considered:

Case i. $u_{1}=u_{2}>l_{1}=l_{2}=p_{2}^{*}.$ As $u_{2}>l_{2}=p_{2}^{*},$ there exists $\xi \in (p_{2}^{*},u_{2})$ such that $F_{2}(\xi )<1.$ Seller 1’s expected profit from posting a price $p\le \xi$ is

$$\begin{aligned} V_{1}(t)= & {} q\delta V_{1}(t)+(1-q)\left[ \phantom {\left. +(F_{2}(p)-F_{2}(p-))\frac{1}{2}(\delta V(\overline{p} ,1)+p)+(1-F_{2}(p))p\right] .}F_{2}(p-)\delta V(\overline{p} ,1)\right. \nonumber \\&\left. +(F_{2}(p)-F_{2}(p-))\frac{1}{2}(\delta V(\overline{p} ,1)+p)+(1-F_{2}(p))p\right] . \end{aligned}$$

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It can be readily seen that $V_{1}(t)=p_{2}^{*}=\delta V(\overline{p},1)$ when $p=p_{2}^{*},$ and $\lim _{p\rightarrow p_{2}^{*}}V_{1}(t)=p_{2}^{*}=\delta V(\overline{p},1).$ On the other hand, the expected profit from posting a price $p=\xi$ is

$$\begin{aligned} V_{1}(t)= & {} q\delta V_{1}(t)+(1-q)\left[ \phantom {\left. -F_{2}(\xi -))\left( \frac{1}{2}\delta V(\overline{p},1)+\frac{1}{2 }\xi \right) +(1-F_{2}(\xi ))\xi \right] }F_{2}(\xi -)\delta V(\overline{p} ,1)+(F_{2}(\xi )\right. \nonumber \\&\left. -F_{2}(\xi -))\left( \frac{1}{2}\delta V(\overline{p},1)+\frac{1}{2 }\xi \right) +(1-F_{2}(\xi ))\xi \right] \nonumber \\> & {} q\delta V_{1}(t)+(1-q)\left[ F_{2}(\xi -)\delta V(\overline{p},1)\right. \nonumber \\&\left. +\,(F_{2}(\xi )-F_{2}(\xi -))\delta V(\overline{p},1)+(1-F_{2}(\xi ))\delta V(\overline{p} ,1)\right] \nonumber \\= & {} \delta V(\overline{p},1). \end{aligned}$$

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Hence, we can find an $\epsilon >0$ such that seller 1’s expected profit at $p=\xi$ is strictly higher than that from posting any price $p\in [p_{2}^{*},p_{2}^{*}+\epsilon ).$ Accordingly, given seller 2’s strategy $F_{2}(\bullet ),$ seller 1 will never choose any $p<p_{2}^{*}+\epsilon$. Therefore $\sup \{p:F_{1}(p)=0\}\ge p_{2}^{*}+\epsilon >p_{2}^{*},$ contradicting the premise that $l_{1}=p_{2}^{*}.$

Case ii. $u_{1}=u_{2}\ge l_{1}=l_{2}>p_{2}^{*}.$ If $F_{2}(u_{2}-)=1,$ then there exists $p_{2}^{*}<\vartheta <u_{2}$ such that $F_{2}(p-)>1-\varepsilon$ $\forall p\in (\vartheta ,u_{2}],$ where $\varepsilon =(l_{1}-p_{2}^{*})/(\delta V(\overline{p},1)/2+3u_{2}/2).$ Then seller 1’s expected profit from posting any price $p\in (\vartheta ,u_{2}]$ is

$$\begin{aligned} V_{1}(t)= & {} q\delta V_{1}(t)+(1-q)\left[ \phantom {\left. -F_{2}(p-))\frac{1}{2}(\delta V(\overline{p} ,1)+p)+(1-F_{2}(p))p\right] }F_{2}(p-)\delta V(\overline{p} ,1)+(F_{2}(p)\right. \nonumber \\&\left. -F_{2}(p-))\frac{1}{2}(\delta V(\overline{p} ,1)+p)+(1-F_{2}(p))p\right] \nonumber \\< & {} q\delta V_{1}(t)+(1-q)[F_{2}(p-)\delta V(\overline{p},1)+\frac{ \varepsilon }{2}(\delta V(\overline{p},1)+p)+\varepsilon p] \nonumber \\\le & {} q\delta V_{1}(t)+(1-q)\left[ \delta V(\overline{p},1)+\varepsilon \left( \frac{1}{ 2}\delta V(\overline{p},1)+\frac{3}{2}p\right) \right] \nonumber \\\le & {} q\delta V_{1}(t)+(1-q)\left[ \delta V(\overline{p},1)+\varepsilon \left( \frac{1}{ 2}\delta V(\overline{p},1)+\frac{3}{2}u_{2}\right) \right] \nonumber \\= & {} q\delta V_{1}(t)+(1-q)l_{1}. \end{aligned}$$

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By choosing the price sufficiently close to $l_{1},$ seller 1’s expected profit can be sufficiently close to $q\delta V_{1}(t)+(1-q)l_{1},$ the last term in the equation above. Therefore seller 1 will never choose a price above $\vartheta$ with positive probability,  which implies that $\inf \{p:F_{1}(p)=1\}\le \vartheta <u_{2}=u_{1},$ a contradiction. Hence in equilibrium we must have $F_{2}(u_{2}-)<1$. Since $u_{2}=\inf \{p:F_{2}(p)=1\},$ $F_{2}(u_{2})-F_{2}(u_{2}-)>0.$ It is then easy to check that seller 1 will never set the price at $u_{1},$ as he can always get higher profit by cutting down the price slightly. Since $u_{1}=\inf \{p:F_{1}(p)=1\},$ this implies that $F_{1}(u_{1}-)=1.$ But, by the same token, $F_{1}(u_{1}-)=1$ also cannot be sustained in any equilibrium. Thus a necessary condition in an equilibrium is that $u_{1}=u_{2}=l_{1}=l_{2}=p_{2}^{*},$ and it is easy to verify that it is indeed an equilibrium. $\square$

Proof of Proposition  2

Since most of the reasoning are parallel to the proof in Proposition 1, here we just give a sketch of the proof. When $N=2$, the conclusion follows from Proposition 1. The proof proceeds by induction. Suppose the Proposition holds in the selling game $\Gamma (N-1,q).$ For the selling game $\Gamma (N,q),$ pick any period t with $S_{t}=N$ and WLOG let $l_{1t}\le l_{2t}\le \cdots \le l_{Nt}.$ Following the same line of reasoning in Claim 1 of Proposition 1, it can be readily verified that $l_{1t}=l_{2t}\ge p_{N}^{*}$ and (a) is established. For part (b), let us define $V_{i}(t)$ to be the present discount value of expected profit to the seller i at period t, and $F_{-i}(p)$ to be the joint probability that all sellers except i set the price less or equal to p. Clearly $F_{-i}(p)$ is right continuous. Now we notice that $l_{1t}=l_{2t}>p_{N}^{*}$ cannot be sustained in any equilibrium. If it were indeed the case, then the expected profit for seller 1 from posting the price at $l_{1t}$ would be

$$\begin{aligned} V_{1}(t)=q\delta V_{1}(t)+(1-q)\left[ F_{-1}(l_{1t})\frac{1}{2}(\delta V(\overline{ p},N-1)+l_{1t})+(1-F_{-1}(l_{1t}))l_{1t}\right] . \end{aligned}$$

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Observing that $F_{-1}(l_{1t})>0$ cannot be the case, as it implies that

$$\begin{aligned} V_{1}(t)\rightarrow & {} q\delta V_{1}(t)+(1-q)\left[ F_{-1}(l_{1t})\frac{1}{2} (\delta V(\overline{p},N-1)+l_{1t})+(1-F_{-1}(l_{1t}))l_{1t}\right] \nonumber \\< & {} q\delta V_{1}(t)+(1-q)l_{1t}\text { as }p\searrow l_{1t}\text { }, \end{aligned}$$

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which implies that either a maximizer does not exist for seller 1 or $\sup \{p:F_{1t}(p)=0\}>l_{1t},$ a contradiction. But $F_{-1}(l_{1t})$ also cannot be equal to zero. $F_{-1}(l_{1t})=0$ implies that conditional on demand being one in period t, the expected profit for every seller is at least $l_{it},$ which in turn implies that $u_{1t}=u_{2t}=\cdots =u_{Nt}$ and $F_{it}(u_{it}-)=1$ $\forall i.$ Also by the right continuity of $F_{-1}(p),$ $F_{it}$ must be non-degenerate for some i. It is easy to see that it cannot be supported in any equilibrium. Consequently $l_{1t}=l_{2t}=p_{N}^{ *}$ is the only feasible outcome. Suppose now that none of the (N-1) sellers chooses $p_{N}^{*}$ with probability 1, then $F_{-1}(p_{N}^{*})<1$ and right continuity again implies that in order to maximize the expected profit, seller 1 will set a price strictly greater than $p_{N}^{*},$ violating the condition $l_{1t}=p_{N}^{*}.$ there has at least one seller, say j, who sets the price at $p_{N}^{*}$ with probability 1. From seller j’s perspective, by the same reasoning, there must have another seller who posts the price at $p_{N}^{*}$ with probability 1. Hence (b) is established. (c) follows from (a) and (b) immediately. $\square$

Proof of Proposition  2

The case $S_{t}=1$ is trivial. Pick any period $t\in \mathbb {N}$ with $S_{t}=2.$ It is easy to show that a pure-strategy equilibrium does not exist in the selling game $\Gamma (2,\{q_{i}\}_{i=0}^{\infty })$ when $q_{1}>0$ and $\sum _{i=2}^{\infty }q_{i}>0$. Now we show that a symmetric mixed-strategy equilibrium exists. Let $F(p):[0,\infty )\rightarrow [0,1]$ denote the mixed-strategy played by each seller in period t in an equilibrium, and we define $l=\sup \{p:F(p)=0\}$ and $u=\inf \{p:F(p)=1\}.$ Following the same line of reasoning in the proof of Proposition 1, it can be established that F(p) is continuous (atomless) and strictly increasing on [l, u] (i.e., the support of F(p) is connected). Define $V_{i}(t)$ to be the present discounted value of seller $i^{\prime }s$ expected profit in period t. Apparently we must have $u=\overline{p}.$ Suppose to the contrary that $u<\overline{p},$ then the expected profit for seller 1 from posting a price $\overline{p}$ is greater than that from posting a price u,  i.e.,

$$\begin{aligned} V_{1}(t)=q_{0}\delta V_{1}(t)+q_{1}\delta V(\overline{p},1)+\sum _{j=2}^{ \infty }q_{j}\overline{p}>q_{0}\delta V_{1}(t)+q_{1}\delta V(\overline{p} ,1)+\sum _{j=2}^{\infty }q_{j}u, \end{aligned}$$

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a contradiction. Since each seller is indifferent between posting a price at $u=\overline{p}$ and posting a price at l,  we must have $l=p_{2}^{*}$ according to the definition of $p_{2}^{*}.$ Then $\forall p\in [p_{2}^{*},\overline{p}]$ the following relation holds:

$$\begin{aligned} (1-\delta q_{0})V(\overline{p},2)=q_{1}[F(p)\delta V(\overline{p} ,1)+(1-F(p))p]+\sum _{i=2}^{\infty }q_{i}p\text { } \end{aligned}$$

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F(p) can be solved as:

$$\begin{aligned} F(p)=\left\{ \begin{array}{ll} 1 &{}\quad p>\overline{p} \\ 1-\frac{\sum _{2}^{\infty }q_{i}}{q_{1}}\left[ \frac{\overline{p}-\delta V( \overline{p},1)}{p-\delta V(\overline{p},1)}-1\right] &{}\quad p_{2}^{*}\le p\le \overline{p} \\ 0 &{}\quad p<p_{2}^{*} \end{array} \right. , \end{aligned}$$

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which is exactly equal to $W_{2}(p).$ $\square$

Proof of Proposition  5

Pick any period $t\in \mathbb {N}$ with $S_{t}=n.$ Let $F_{n}(p):[0,\infty )\rightarrow [0,1]$ denote the mixed-strategy played by each seller in period t in an equilibrium. We divide the proof into two steps:

Step 1: No price dispersion when sellers face low demand uncertainty

First we provide a necessary and sufficient condition for the unique existence of symmetric pure-strategy equilibrium. Suppose $(p^{*},\ldots ,p^{*})$ constitutes a symmetric pure-strategy equilibrium in period t. Define $\widehat{V}(k)$ to be the present discounted value of a seller’s expected profit when there are k sellers in the market. Clearly $\widehat{V}(1)=V(\overline{p},1),$ and $\widehat{V}(2)=V(\overline{p},2)$ from Propositions 2 and 4. By induction, we claim that $\widehat{V}(k)=V( \overline{p},k)$ for every $k\in \mathbb {N}$ in equilibrium, and the claim is verified along the proof. Given $(p^{*},\ldots ,p^{*})$ is an equilibrium, the following two conditions should be satisfied:

(a) No seller has an incentive to slightly undercut its competitors. This condition requires that

$$\begin{aligned}&q_{0}\delta V(\overline{p},n)+q_{1}\left( \frac{1}{n}p^{*}+\left( 1-\frac{1}{n} \right) \delta V(\overline{p},n-1)\right) \nonumber \\&\qquad +\cdots +q_{n-1}\left( \frac{n-1}{n}p^{*}+\frac{1}{n}\delta V(\overline{p} ,1)\right) +\sum _{i=n}^{\infty }q_{i}p^{*}\\&\ge q_{0}\delta V(\overline{p} ,n)+\sum _{i=1}^{\infty }q_{i}p^{*} \nonumber \\&\quad \Longrightarrow p^{*} \le \frac{\sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n} )\delta V(\overline{p},n-i)}{\sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n})} \nonumber \end{aligned}$$

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(b) No seller has an incentive to raise its price to $\overline{p}.$ This condition requires that

$$\begin{aligned}&q_{0}\delta V(\overline{p},n)+q_{1}\left( \frac{1}{n}p^{*}+\left( 1-\frac{1}{n} \right) \delta V(\overline{p},n-1)\right) \nonumber \\&\qquad +\cdots +q_{n-1}\left( \frac{n-1}{n}p^{*}+\frac{1}{n}\delta V(\overline{p} ,1)\right) +\sum _{i=n}^{\infty }q_{i}p^{*}\\&\quad \ge q_{0}\delta V(\overline{p} ,n)+\sum _{i=1}^{n-1}q_{i}\delta V(\overline{p},n-i)+\sum _{i=n}^{\infty }q_{i} \overline{p} \nonumber \\&\quad \Longrightarrow p^{*} \ge \frac{\sum _{i=1}^{n-1}q_{i}\frac{i}{n} \delta V(\overline{p},n-i)+\sum _{i=n}^{\infty }q_{i}\overline{p}}{ \sum _{i=1}^{n-1}q_{i}\frac{i}{n}+\sum _{i=n}^{\infty }q_{i}} \nonumber \end{aligned}$$

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Therefore, $(p^{*},\ldots ,p^{*})$ constitutes an equilibrium if and only if the following inequality holds:

$$\begin{aligned} \frac{\sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n})\delta V(\overline{p},n-i)}{ \sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n})}\ge \frac{\sum _{i=1}^{n-1}q_{i}\frac{i }{n}\delta V(\overline{p},n-i)+\sum _{i=n}^{\infty }q_{i}\overline{p}}{ \sum _{i=1}^{n-1}q_{i}\frac{i}{n}+\sum _{i=n}^{\infty }q_{i}}. \end{aligned}$$

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Let us consider four subcases:

Case 1. $\Lambda (n,\{q_{i}\}_{i=0}^{\infty })=1$ (one non-zero demand state). In this case $\underline{i}=\overline{i}$ is the only $i\in \{1,\ldots ,n-1,n+\}$ with $q_{i}>0$.¹⁴ It can be readily seen that the inequality holds and $(p^{*},\ldots , p^{*})$ constitutes an equilibrium with $p^{*}=\left\{ \begin{array}{cc} \delta V(\overline{p},n-\overline{i}) &{} \overline{i}\ne n+ \\ \overline{p} &{} \overline{i}=n+ \end{array} \right. =\eta _{n}.$

Moreover, in equilibrium each seller’s expected profit is $V(\overline{p} ,n),$ so our claim $\widehat{V}(n)=V(\overline{p},n)$ is verified.

Case 2. $\Lambda (n,\{q_{i}\}_{i=0}^{\infty })\ge 2$ with $\overline{i}=n+$ (i.e., $\sum _{i=n}^{\infty }q_{i}=q_{n+}>0).$ As $\delta V( \overline{p},n-1)\le \delta V(\overline{p},n-2)\le \cdots \le \delta V( \overline{p},1)<\overline{p}$, we have

$$\begin{aligned}&\frac{\sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n})\delta V(\overline{p},n-i)}{ \sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n})}\le \frac{\sum _{i=1}^{n-1}q_{i}\frac{i }{n}\delta V(\overline{p},n-i)}{\sum _{i=1}^{n-1}q_{i}\frac{i}{n}}\\&\quad <\frac{ \sum _{i=1}^{n-1}q_{i}\frac{i}{n}\delta V(\overline{p},n-i)+\sum _{i=n}^{ \infty }q_{i}\overline{p}}{\sum _{i=1}^{n-1}q_{i}\frac{i}{n} +\sum _{i=n}^{\infty }q_{i}}. \end{aligned}$$

Hence the inequality does not hold and there exists no symmetric pure-strategy equilibrium.

Case 3. $\Lambda (n,\{q_{i}\}_{i=0}^{\infty })\ge 2$, $\overline{i}\ne n+$ and $V(\overline{p},n-\overline{i})=V(\overline{p},n- \underline{i}).$ By definition, $V(\overline{p},n-\overline{i})\ge V( \overline{p},j)\ge V(\overline{p},n-\underline{i})$ for every $n-\overline{i }<j<n-\underline{i}.$ $V(\overline{p},n-\overline{i})=V(\overline{p},n- \underline{i})$ implies $V(\overline{p},n-\overline{i})=V(\overline{p},j)$ for every $n-\overline{i}<j<n-\underline{i}.$ Accordingly,

$$\begin{aligned}&\frac{\sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n})\delta V(\overline{p},n-i)}{ \sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n})} =\frac{\sum _{i=\underline{i}}^{ \overline{i}}q_{i}(1-\frac{i}{n})\delta V(\overline{p},n-i)}{\sum _{i= \underline{i}}^{\overline{i}}q_{i}(1-\frac{i}{n})}\\&\quad =\delta V(\overline{p},n- \underline{i})=\frac{\sum _{i=\underline{i}}^{\overline{i}}q_{i}\frac{i}{n} \delta V(\overline{p},n-i)}{\sum _{i=\underline{i}}^{\overline{i}}q_{i}\frac{i }{n}} \\&\quad =\frac{\sum _{i=1}^{n-1}q_{i}\frac{i}{n}\delta V(\overline{p},n-i)}{ \sum _{i=1}^{n-1}q_{i}\frac{i}{n}}=\frac{\sum _{i=1}^{n-1}q_{i}\frac{i}{n} \delta V(\overline{p},n-i)+\sum _{i=n}^{\infty }q_{i}\overline{p}}{ \sum _{i=1}^{n-1}q_{i}\frac{i}{n}+\sum _{i=n}^{\infty }q_{i}} \end{aligned}$$

Therefore the inequality holds and $(p^{*},\ldots ,p^{*})$ constitutes an equilibrium with $p^{*}=\delta V(\overline{p},n-\underline{i})=V( \overline{p},n-\overline{i}).$ In equilibrium each seller’s expected profit is $V(\overline{p},n),$ and our claim $\widehat{V}(n)=V(\overline{p},n)$ is verified.

Case 4. $\Lambda (n,\{q_{i}\}_{i=0}^{\infty })\ge 2$, $\overline{i}\ne n+$ and $V(\overline{p},n-\overline{i})>V(\overline{p},n- \underline{i}).$ As $\delta V(\overline{p},n-1)\le \delta V(\overline{p} ,n-2)\le \cdots \le \delta V(\overline{p},1),$ $1-\frac{i}{n}$ is decreasing in i and $\frac{i}{n}$ is increasing in i, we observe that

$$\begin{aligned} \frac{\sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n})\delta V(\overline{p},n-i)}{ \sum _{i=1}^{n-1}q_{i}(1-\frac{i}{n})}\le \frac{\sum _{i=1}^{n-1}q_{i}\frac{i }{n}\delta V(\overline{p},n-i)}{\sum _{i=1}^{n-1}q_{i}\frac{i}{n}}. \end{aligned}$$

Moreover, this inequality holds strictly whenever there are $i,j\in \{1,\ldots ,n-1\}$ such that $q_{i}>0,q_{j}>0,$ and $V(\overline{p},n-i)\ne V( \overline{p},n-j).$ Here we have $q_{\overline{i}}>0,q_{\underline{i}}>0$ and $V(\overline{p},n-\overline{i})>V(\overline{p},n-\underline{i}),$ and hence the above inequality holds strictly. This directly implies that the equilibrium condition does not hold, and hence there exists no symmetric pure-strategy equilibrium under this case.

To summarize, we have shown that there exists a unique symmetric pure-strategy equilibrium (i.e., $F_{n}$ is degenerate) if and only if the degree of demand uncertainty is low. When the degree of demand uncertainty is high, $F_{n}$ is non-degenerate. In the next step we show that $F_{n}=W_{n}$ (defined in Eq. (14)).

Step 2. Price dispersion when sellers face high demand uncertainty

We prove it by induction. $F_{2}=W_{2}$ in a high demand state from Proposition 4, and the expected profit for each seller is $V(\overline{p},2)$ in equilibrium. Suppose each seller’s expected profit is $V(\overline{p},i)$ in any period t with i sellers remaining in the market, $i=2,\ldots ,n-1.$ We show that in a high demand state $F_{n}=W_{n}$ constitutes a unique equilibrium strategy in period t with $S_{t}=n>\underline{i}.$ First we establish a few lemmas to show that $G_{n}$ (defined in Eq. (13)) is strictly increasing when the degree of demand uncertainty is high, and hence $W_{n}$ is a non-degenerate distribution.

Lemma 1

Let $\Phi _{n}(x): \mathbb {R} \rightarrow \mathbb {R}$ be of the form

$$\begin{aligned} \Phi _{n}(x)=\frac{\alpha _{1}(x)C_{1}+\alpha _{2}(x)C_{2}+\cdots +\alpha _{n}(x)C_{n}}{\alpha _{1}(x)+\alpha _{2}(x)+\cdots +\alpha _{n}(x)}, \end{aligned}$$

where $C_{1}\le C_{2}\le \cdots \le C_{n}$, $\alpha _{i}(x)\ge 0$ is differentiable for $i=1,2,\ldots ,n$, and $\sum _{i=1}^{n}\alpha _{i}(x)\ne 0$ $\forall x\in \mathbb {R} .$ Then $\frac{\partial }{\partial x}\Phi _{n}(x)>0$ if $\alpha _{j}(x)\frac{ \partial }{\partial x}[\sum _{i=j+1}^{n}\alpha _{i}(x)]-\sum _{i=j+1}^{n}\alpha _{i}(x)\frac{\partial }{\partial x}\alpha _{j}(x)\ge 0$ for $j=1,\ldots ,n-1,$ and this holds with strict inequality for some $j^{*}$ and $C_{j^{*}}<\frac{\sum _{i=j^{*}+1}^{n}\alpha _{i}(x)C_{i}}{\sum _{i=j^{*}+1}^{n}\alpha _{i}(x)}$.

Proof

We prove it by induction. The result is trivially true for $n=2$. When $n=3$, we can rewrite $\Phi _{3}(x)$ as

$$\begin{aligned} \Phi _{3}(x)= & {} \frac{\alpha _{1}(x)C_{1}+\alpha _{2}(x)C_{2}+\alpha _{3}(x)C_{3}}{\alpha _{1}(x)+\alpha _{2}(x)+\alpha _{3}(x)}\nonumber \\= & {} \frac{\alpha _{1}(x)C_{1}+[\alpha _{2}(x)+\alpha _{3}(x)]\{\frac{\alpha _{2}(x)C_{2}+\alpha _{3}(x)C_{3}}{\alpha _{2}(x)+\alpha _{3}(x)}\}}{\alpha _{1}(x)+[\alpha _{2}(x)+\alpha _{3}(x)]} \end{aligned}$$

(34)

First taking $\frac{\alpha _{2}(x)C_{2}+\alpha _{3}(x)C_{3}}{\alpha _{2}(x)+\alpha _{3}(x)}$ as constant and applying the $n=2$ result, $\frac{ \partial }{\partial x}\Phi _{3}(x)>0$ if $\alpha _{1}(x)\frac{\partial }{ \partial x}[\alpha _{2}(x)+\alpha _{3}(x)]-[\alpha _{2}(x)+\alpha _{3}(x)] \frac{\partial }{\partial x}\alpha _{1}(x)>0$ and $C_{1}<\frac{\alpha _{2}(x)C_{2}+\alpha _{3}(x)C_{3}}{\alpha _{2}(x)+\alpha _{3}(x)}$ (condition (i)). Second, applying the $n=2$ result on $\frac{\alpha _{2}(x)C_{2}+\alpha _{3}(x)C_{3}}{\alpha _{2}(x)+\alpha _{3}(x)}$ we get $\frac{\partial }{ \partial x}\frac{\alpha _{2}(x)C_{2}+\alpha _{3}(x)C_{3}}{\alpha _{2}(x)+\alpha _{3}(x)}>0$ if $\alpha _{2}(x)\frac{\partial }{\partial x} \alpha _{3}(x)-\alpha _{3}(x)\frac{\partial }{\partial x}\alpha _{2}(x)>0$ and $C_{2}<C_{3}$ (condition (ii)). Then it can be readily seen that $\Phi _{3}(x)>0$ when either (i) or (ii) holds, and the other condition holds weakly. Hence the $n=3$ case is established. Inductively applying the same argument for any $n\in \mathbb {N} ,$ we get the conditions stated in the lemma. $\square$

Lemma 2

$\frac{\partial }{\partial x}Z_{k,n}(x)=-(k+1)\left( {\begin{array}{c}n-1\\ k+1\end{array}}\right) (1-x)^{n-2-k}x^{k}<0 \forall x\in (0,1).$

Proof

$$\begin{aligned} \frac{\partial }{\partial x}Z_{k,n}(x)= & {} \frac{\partial }{\partial x} \sum _{i=0}^{k}\left( {\begin{array}{c}n-1\\ i\end{array}}\right) (1-x)^{n-1-i}x^{i} \nonumber \\= & {} \sum _{i=0}^{k}\left[ \phantom {\left. +i\left( {\begin{array}{c}n-1\\ i\end{array}}\right) (1-x)^{n-1-i}x^{i-1}\right] }-(n-1-i)\left( {\begin{array}{c}n-1\\ i\end{array}}\right) (1-x)^{n-2-i}x^{i}\right. \nonumber \\&\left. +i\left( {\begin{array}{c}n-1\\ i\end{array}}\right) (1-x)^{n-1-i}x^{i-1}\right] \nonumber \\= & {} \sum _{i=0}^{k}\left[ -(i+1)\left( {\begin{array}{c}n-1\\ i+1\end{array}}\right) (1-x)^{n-2-i}x^{i}+i\left( {\begin{array}{c}n-1\\ i\end{array}}\right) (1-x)^{n-1-i}x^{i-1}\right] \nonumber \\= & {} -(k+1)\left( {\begin{array}{c}n-1\\ k+1\end{array}}\right) (1-x)^{n-2-k}x^{k}. \end{aligned}$$

(35)

$\square$

Lemma 3

Given $q_{i}>0$ and $q_{j}>0$ for some $i,j\in \{1,\ldots ,n-1,n+\},$ $\frac{ \partial }{\partial x}G_{n}(x)>0$ $\forall x\in (0,1).$

Proof

We observe that $\delta V(\overline{p},n-1)\le \delta V(\overline{p} ,n-2)\le \cdots \le \delta V(\overline{p},1)<\overline{p},$ $\{q_{i}Z_{i-1,n}(x)\}_{i=1}^{n-1}\ge 0,$ $\sum _{i=n}^{\infty }q_{i}\ge 0,$ and $\sum _{i=1}^{n-1}q_{i}Z_{i-1,n}(x)+\sum _{i=n}^{\infty }q_{i}>0$ $\forall x\in (0,1).$ Hence, $G_{n}(x)$ meets the functional form of $\Phi _{n}(x)$ specified in Lemma 1. By Lemma 1, $\frac{\partial }{\partial x}G_{n}(x)>0$ $\forall x\in (0,1)$ if the following conditions are satisfied:

$$\begin{aligned}&q_{n-1}Z_{n-2,n}(x)\frac{\partial }{\partial x}\left[ \sum _{i=n}^{\infty }q_{i}\right] -\left[ \sum _{i=n}^{\infty }q_{i}\right] \frac{\partial }{\partial x} [q_{n-1}Z_{n-2,n}(x)]\ge 0 \end{aligned}$$

(36)

$$\begin{aligned}&\Psi _{j}(x)= \begin{array}{l} q_{n-j+1}Z_{n-j,n}(x)\frac{\partial }{\partial x} \left[ \sum _{i=1}^{j-2}q_{n-j+1+i}Z_{n-j+i,n}(x)+\sum _{i=n}^{\infty }q_{i}\right] \\ -[\sum _{i=1}^{j-2}q_{n-j+1+i}Z_{n-j+i,n}(x)+\sum _{i=n}^{\infty }q_{i}] \frac{\partial }{\partial x}[q_{n-j+1}Z_{n-j,n}(x)] \end{array}\nonumber \\&\qquad \qquad \ge 0,\text { }j=3,\ldots ,n. \end{aligned}$$

(37)

Moreover, one of the above conditions holds with strict inequality for some $j^{*}$ and

$$\begin{aligned} \delta V(\overline{p},j^{*}-1)<\frac{\sum _{i=1}^{j^{*}-2}q_{n-j^{*}+1+i}Z_{n-j^{*}+i,n}(x)\delta V(\overline{p},j^{*}-1-i)+\sum _{i=n}^{\infty }q_{i}\overline{p}}{\sum _{i=1}^{j^{*}-2}q_{n-j^{*}+1+i}Z_{n-j^{*}+i,n}(x)+\sum _{i=n}^{\infty }q_{i}}. \end{aligned}$$

(38)

By Lemma 2, we have

$$\begin{aligned}&q_{n-1}Z_{n-2,n}(x)\frac{\partial }{\partial x}\left[ \sum _{i=n}^{\infty }q_{i}\right] -\left[ \sum _{i=n}^{\infty }q_{i}\right] \frac{\partial }{\partial x} [q_{n-1}Z_{n-2,n}(x)]\nonumber \\&\quad =\left[ \sum _{i=n}^{\infty }q_{i}\right] q_{n-1}(n-1)x^{n-2}\ge 0. \end{aligned}$$

(39)

We observe that

$$\begin{aligned} \Psi _{j}(x)= & {} q_{n-j+1}Z_{n-j,n}(x)\frac{\partial }{\partial x} \left[ \sum _{i=1}^{j-2}q_{n-j+1+i}Z_{n-j+i,n}(x)+\sum _{i=n}^{\infty }q_{i}\right] \nonumber \\&-\left[ \sum _{i=1}^{j-2}q_{n-j+1+i}Z_{n-j+i,n}(x)+\sum _{i=n}^{\infty }q_{i}\right] \frac{\partial }{\partial x}[q_{n-j+1}Z_{n-j,n}(x)] \nonumber \\\ge & {} \Gamma _{j}(x)\equiv q_{n-j+1}Z_{n-j,n}(x)\frac{\partial }{\partial x} \left[ \sum _{i=1}^{j-2}q_{n-j+1+i}Z_{n-j+i,n}(x)\right] \nonumber \\&-\left[ \sum _{i=1}^{j-2}q_{n-j+1+i}Z_{n-j+i,n}(x)\right] \frac{\partial }{\partial x} [q_{n-j+1}Z_{n-j,n}(x)] \end{aligned}$$

(40)

After simplification, we get:

$$\begin{aligned} \Gamma _{j}(x)= & {} q_{n-j+1}\sum _{i=1}^{j-2}\sum _{k=n-j+1}^{n-j+i} \sum _{l=0}^{n-j}q_{n-j+1+i}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \\&\left( {\begin{array}{c}n-1\\ l\end{array}}\right) [(1-x)^{2n-2-l-k}x^{l+k}](k-l)\frac{\partial }{\partial x}\log \frac{x}{1-x} \ge 0 \end{aligned}$$

Therefore, conditions (36) and (37) are satisfied. Next we show that at least one condition holds with strict inequality when the degree of demand uncertainty is high. Recall that a selling game $\Gamma (n,\{q_{i}\}_{i=0}^{\infty })$ is said to have a high degree of demand uncertainty if (i) $\Lambda (n,\{q_{i}\}_{i=0}^{\infty })\ge 2$ and $\overline{i}=n+,$ or (ii) $\Lambda (n,\{q_{i}\}_{i=0}^{\infty })\ge 2,$ $\overline{i}\ne n+$ and $V(\overline{p},n-\overline{i})>V(\overline{p},n- \underline{i}).$ Consider three subcases:

Case 1. $\sum _{i=n}^{\infty }q_{i}>0$ ($\overline{i}=n+)$ and $q_{n-1}>0.$ Since $\delta V(\overline{p},1)<\overline{p}$ and condition (39) holds with strict inequality when $\sum _{i=n}^{\infty }q_{i}>0$ and $q_{n-1}>0,$ by Lemma 1 $\frac{\partial }{\partial x}G_{n}(x)>0$ $\forall x\in (0,1).$

Case 2. $\sum _{i=n}^{\infty }q_{i}>0$ ($\overline{i}=n+)$ and $q_{n-1}=0.$ Let $j^{*}=n+1-\underline{i}.$ As $\sum _{i=n}^{\infty }q_{i}>0$ and $\delta V(\overline{p},j^{*}-1)\le \cdots \le \delta V( \overline{p},1)<\overline{p},$ condition (38) holds. Moreover, by Lemma 2 we have

$$\begin{aligned} \Psi _{j^{*}}(x)= & {} \Gamma _{j^{*}}(x)-\sum _{i=n}^{\infty }q_{i}\frac{ \partial }{\partial x}[q_{n-j^{*}+1}Z_{n-j^{*},n}(x)]\\= & {} \Gamma _{j^{*}}(x)-q_{\underline{i}}\sum _{i=n}^{\infty }q_{i}\frac{\partial }{ \partial x}Z_{n-j^{*},n}(x)>\Gamma _{j^{*}}(x)\ge 0. \end{aligned}$$

Thus, by Lemma 1, $\frac{\partial }{\partial x}G_{n}(x)>0$ $\forall x\in (0,1).$

Case 3. $\sum _{i=n}^{\infty }q_{i}=0$ ( $\overline{i}\ne n+)$ and $V(\overline{p},n-\overline{i})>V(\overline{p},n-\underline{i}).$ Let $j^{*}=n+1-\underline{i}$. Since $q_{n-j^{*}+1}=q_{\underline{i} }>0$ and $q_{\overline{i}}>0$ with $\overline{i}>\underline{i}$,

$$\begin{aligned} \Psi _{j^{*}}(x)= & {} \Gamma _{j^{*}}(x) \\= & {} q_{\underline{i}}\sum _{i=1}^{j^{*}-2}q_{n-j^{*}+1+i}\sum _{k=n-j^{*}+1}^{n-j^{*}+i}\sum _{l=0}^{n-j^{*}}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \\&\left( {\begin{array}{c}n-1\\ l\end{array}}\right) [(1-x)^{2n-2-l-k}x^{l+k}](k-l)\frac{\partial }{\partial x}\log \frac{x}{1-x} \\\ge & {} q_{\underline{i}}q_{\overline{i}}\sum _{k=n-j^{*}+1}^{n-j^{*}+i}\sum _{l=0}^{n-j^{*}}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \\&\left( {\begin{array}{c}n-1\\ l\end{array}}\right) [(1-x)^{2n-2-l-k}x^{l+k}](k-l)\frac{\partial }{\partial x}\log \frac{x}{1-x} >0. \end{aligned}$$

Next we observe that $\delta V(\overline{p},j^{*}-1)=\delta V(\overline{p },n-\underline{i})<\delta V(\overline{p},j^{*}-1-\overline{i}+\underline{ i})=\delta V(\overline{p},n-\overline{i}),$ which in turn implies

$$\begin{aligned} \delta V(\overline{p},j^{*}-1)<\frac{\sum _{i=1}^{j^{*}-2}q_{n-j^{*}+1+i}Z_{n-j^{*}+i,n}(x)\delta V(\overline{p},j^{*}-1-i)+\sum _{i=n}^{\infty }q_{i}\overline{p}}{\sum _{i=1}^{j^{*}-2}q_{n-j^{*}+1+i}Z_{n-j^{*}+i,n}(x)+\sum _{i=n}^{\infty }q_{i}}. \end{aligned}$$

Again, by Lemma 1, $\frac{\partial }{\partial x}G_{n}(x)>0$ $\forall x\in (0,1)$ in this case, and the proof of this lemma is complete. $\square$

Lemma 4

$W_{n}(p)$ is a probability distribution with $W_{n}(\eta _{n})=\lim _{x\rightarrow \eta _{n}}G_{n}^{-1}(x)=1$ and $W_{n}(p_{n}^{*})=0.$

Proof

It can be readily verified that $W_{n}(p_{n}^{*})=G_{n}^{-1}(p_{n}^{*})=0$ and $W_{n}(\eta _{n})=\lim _{x\rightarrow \eta _{n}}G_{n}^{-1}(x)=1.$ Since $G_{n}(p)$ is strictly increasing in [0, 1) , $W_{n}(p)=G_{n}^{-1}(p)$ is well defined and strictly increasing in $[p_{n}^{*},\eta _{n}].$

Now we are in the position to prove that $F_{n}=W_{n}.$ As each seller plays $F_{n}$ constitutes an equilibrium, no one will deviate if the other $n-1$ sellers play mixed-strategies according to $F_{n}$ when there are n sellers in the market in period t. It is equivalent to saying that setting any price between $[l_{n},u_{n}]$ gives sellers the same expected profits, $i.e.,\forall p\in [l_{n},u_{n}],$

$$\begin{aligned} V(\overline{p},n)= & {} \frac{1}{1-\delta q_{0}}\left[ \sum _{i=1}^{n-1}q_{i}\delta V( \overline{p},n-i)+\sum _{i=n}^{\infty }q_{i}\overline{p}\right] \nonumber \\= & {} \frac{1}{1-\delta q_{0}}\left\{ \sum _{i=1}^{n-1}q_{i}[Z_{i-1}(F_{n}(p))p+(1-Z_{i-1}(F_{n}(p)))\delta V( \overline{p},n-i)]\right. \nonumber \\&\left. +\sum _{i=n}^{\infty }q_{i}p\right\} , \end{aligned}$$

(41)

or

$$\begin{aligned} \sum _{i=n}^{\infty }q_{i}\overline{p} =\sum _{i=1}^{n-1}q_{i}Z_{i-1}(F_{n}(p))[p-\delta V(\overline{p} ,n-i)]+\sum _{i=n}^{\infty }q_{i}p. \end{aligned}$$

(42)

Therefore,

$$\begin{aligned} p=\frac{\sum _{i=1}^{n-1}q_{i}Z_{i-1}(F_{n}(p))\delta V(\overline{p} ,n-i)+\sum _{i=n}^{\infty }q_{i}\overline{p}}{ \sum _{i=1}^{n-1}q_{i}Z_{i-1}(F_{n}(p))+\sum _{i=n}^{\infty }q_{i}}. \end{aligned}$$

We already knew that $W_{n}(p)=G_{n}^{-1}(p)$ is the unique solution to this equation in $[p_{n}^{*},\eta _{n}]$. Moreover, in equilibrium we must have $l_{n}=p_{n}^{*}$ and $u_{n}=\eta _{n}.$ Hence $F_{n}=W_{n}$ and each seller’s expected profit is $V(\overline{p},n)$ in period t. Q.E.D.

Proof of the Remark

Suppose $k^{*}=1.$ In this case, $n-\overline{ i}\le \overline{i}.$ Consider three subcases:

• (i) If $n-\underline{i}>\overline{i},$ then

  $$\begin{aligned} V(\overline{p},n-\overline{i})= & {} \frac{\sum _{j=\underline{i}}^{n-\overline{i} -1}q_{j}\delta V(\overline{p},n-\overline{i}-j)+\sum _{j=n-\overline{i}}^{ \overline{i}}q_{j}\overline{p}}{1-\delta q_{0}}\\> & {} \frac{\sum _{j=\underline{i} }^{\overline{i}}q_{j}\delta V(\overline{p},n-\underline{i}-j)}{1-\delta q_{0} }=V(\overline{p},n-\underline{i}). \end{aligned}$$

• (ii) If $n-\overline{i}\le \underline{i}<n-\underline{i}\le \overline{i},$ then

  $$\begin{aligned} V(\overline{p},n-\overline{i})= & {} \frac{(1-q_{0})\overline{p}}{1-\delta q_{0}}> \frac{\sum _{j=\underline{i}}^{n-\underline{i}-1}q_{j}\delta V(\overline{p},n- \underline{i}-j)+\sum _{j=n-\underline{i}}^{\overline{i}}q_{j}\overline{p}}{ 1-\delta q_{0}}\\= & {} V(\overline{p},n-\underline{i}). \end{aligned}$$

• (iv) If $n-\overline{i}<n-\underline{i}\le \underline{i},$ then

  $$\begin{aligned} V(\overline{p},n-\overline{i})=\frac{(1-q_{0})\overline{p}}{1-\delta q_{0}} =V(\overline{p},n-\underline{i}). \end{aligned}$$

  Thus, $V(\overline{p},n-\overline{i})>V(\overline{p},n-\underline{i})$ if and only if $n>(k^{*}+1)\underline{i}=2\underline{i}.$

Now suppose $\Gamma (n,\{q_{i}\}_{i=0}^{\infty })$ is a selling game with $k^{*}=m>1.$ First we show that for every $h\in \{1,\ldots ,m-1\}, V( \overline{p},n-h\overline{i})>V(\overline{p},n-h\underline{i})$ if and only if $V(\overline{p},n-(h+1)\overline{i})>V(\overline{p},n-(h+1)\underline{i} ).$ By definition,

$$\begin{aligned} V(\overline{p},n-h\overline{i})= & {} \frac{\sum _{j=\underline{i}}^{\overline{i} }q_{j}\delta V(\overline{p},n-h\overline{i}-j)}{1-\delta q_{0}}\\\ge & {} V( \overline{p},n-h\underline{i})=\frac{\sum _{j=\underline{i}}^{\overline{i} }q_{j}\delta V(\overline{p},n-h\underline{i}-j)}{1-\delta q_{0}}. \end{aligned}$$

If $V(\overline{p},n-(h+1)\overline{i})=V(\overline{p},n-(h+1)\underline{i} ),$ then by the weak monotonicity of $V(\overline{p},\cdot ),$ $V(\overline{ p},n-j)=V(\overline{p},n-(h+1)\underline{i})$ for $j\in \{(h+1)\underline{i} +1,\ldots ,(h+1)\overline{i}\},$ which in turn implies that $V(\overline{p},n-h \overline{i})=V(\overline{p},n-h\underline{i}).$ On the other hand, as both $q_{\underline{i}}$ and $q_{\overline{i}}$ are positive, $V(\overline{p},n-h \overline{i})=V(\overline{p},n-h\underline{i})$ implies that $V(\overline{p} ,n-h\overline{i}-\underline{i})=$ $V(\overline{p},n-(h+1)\underline{i})$ and $V(\overline{p},n-(h+1)\overline{i})=V(\overline{p},n-h\underline{i}- \overline{i}).$ Since $n-(h+1)\overline{i}<n-h\overline{i}-\underline{i}<n-h \underline{i}-\overline{i}<n-(h+1)\underline{i},$ by the weak monotonicity of $V(\overline{p},\cdot )$ again, $V(\overline{p},n-(h+1)\underline{i})=V( \overline{p},n-(h+1)\overline{i}).$ Thus, $V(\overline{p},n-h\overline{i})>V( \overline{p},n-h\underline{i})$ if and only if $V(\overline{p},n-(h+1) \overline{i})>V(\overline{p},n-(h+1)\underline{i}).$

Since it holds for every $h\in \{1,\ldots ,m-1\},$ by induction, $V(\overline{p} ,n-\overline{i})>V(\overline{p},n-\underline{i})$ if and only if $V( \overline{p},n-m\overline{i})>V(\overline{p},n-m\underline{i}).$ Therefore, the remark is established if we are able to show that $V(\overline{p},n-m \overline{i})>V(\overline{p},n-m\underline{i})$ if and only if $n>(m+1) \underline{i}.$ Note that from the definition of $k^{*},$ $n-m\overline{i }\le \overline{i}.$ Consider three subcases:

• (i) If $n-m\underline{i}>\overline{i},$ then

  $$\begin{aligned} V(\overline{p},n-m\overline{i})= & {} \frac{\sum _{j=\underline{i}}^{n-m\overline{i} -1}q_{j}\delta V(\overline{p},n-m\overline{i}-j)+\sum _{j=n-m\overline{i}}^{ \overline{i}}q_{j}\overline{p}}{1-\delta q_{0}}\\> & {} \frac{\sum _{j=\underline{i} }^{\overline{i}}q_{j}\delta V(\overline{p},n-m\underline{i}-j)}{1-\delta q_{0}}=V(\overline{p},n-m\underline{i}). \end{aligned}$$

• (ii) If $n-m\overline{i}\le \underline{i}<n-m\underline{i}\le \overline{i} ,$ then

  $$\begin{aligned} V(\overline{p},n-m\overline{i})= & {} \frac{(1-q_{0})\overline{p}}{1-\delta q_{0}}\\> & {} \frac{\sum _{j=\underline{i}}^{n-m\underline{i}-1}q_{j}\delta V(\overline{p} ,n-m\underline{i}-j)+\sum _{j=n-m\underline{i}}^{\overline{i}}q_{j}\overline{p }}{1-\delta q_{0}}\\&=V(\overline{p},n-m\underline{i}). \end{aligned}$$

• (iii) If $n-m\overline{i}<n-m\underline{i}\le \underline{i},$ then

  $$\begin{aligned} V(\overline{p},n-m\overline{i})=\frac{(1-q_{0})\overline{p}}{1-\delta q_{0}} =V(\overline{p},n-m\underline{i}). \end{aligned}$$

Accordingly, $V(\overline{p},n-m\overline{i})>V(\overline{p},n-m\underline{i} )$ if and only if $n>(m+1)\underline{i}.$ $\square$